# Thread: Finding a power series

1. ## Finding a power series

Question: Expand $\displaystyle \frac{1}{(1+x)^\frac{1}{4}}$ as a power series and use it to estimate $\displaystyle \frac{1}{1.1^\frac{1}{4}}$ correct to three decimal places.

I am clueless!

2. Originally Posted by Em Yeu Anh
Question: Expand $\displaystyle \frac{1}{(1+x)^\frac{1}{4}}$ as a power series and use it to estimate $\displaystyle \frac{1}{1.1^\frac{1}{4}}$ correct to three decimal places.

I am clueless!
Binomial series for non-integer exponents:

$\displaystyle (1+x)^{\frac{1}{4}}= ..$

Then plug in $\displaystyle x=0.1$, and as the series is alternating the absolute value of the error is less than the absolute value of the first ignored term

CB

3. Originally Posted by Em Yeu Anh
Question: Expand $\displaystyle \frac{1}{(1+x)^\frac{1}{4}}$ as a power series and use it to estimate $\displaystyle \frac{1}{1.1^\frac{1}{4}}$ correct to three decimal places.

I am clueless!
That is $\displaystyle f(x)= (1+ x)^{-1/4}$

$\displaystyle f'(x)= -\frac{1}{4}(1+ x)^{-5/4}$

$\displaystyle f"(x)= \frac{5}{16}(1+ x)^{-9/4}$

$\displaystyle f'''(x)= -\frac{45}{64}(1+ x)^{-13/4}$

The nth derivative is, I believe, $\displaystyle \frac{(-1)^n}{1(3)(9)\cdot\cdot\cdot(4n+1)}(1+ x)^{-(4n+1)/4}$.

Write it as a Taylor's series about x= 1. Since the sign is alternating, the error is never larger than the next term in the series. Expand it until the next term is less than .001.

(Well, yeah, the generalized binomial series is probably easier!)

4. Originally Posted by Em Yeu Anh
Question: Expand $\displaystyle \frac{1}{(1+x)^\frac{1}{4}}$ as a power series and use it to estimate $\displaystyle \frac{1}{1.1^\frac{1}{4}}$ correct to three decimal places.

I am clueless!
Also check out "Binomial Theorem" on Wikipedia. Rising factorial, falling factorial, I'm confused. How about I show you this Mathematica code and then you figure out how I got it and what it means.

Code:
n[858]:=
myFunctions[x_] := 1/(1 + x)^(1/4)
mySeries[x_, n_] :=
Sum[(FactorialPower[-4^(-1), k]/k!)*
x^k, {k, 0, n}]
mySeries[0.1, 20]
myFunctions[0.1]

Out[860]=
0.9764540896763106

Out[861]=
0.9764540896763105